算法问题实战策略 WORDCHAIN

地址  https://algospot.com/judge/problem/read/WORDCHAIN

解答:

1 书上的解法是制作有向图 然后查找欧拉回路  代码实现稍后

 假设一定存在欧拉路径的做法

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

vector<vector<int>> adj;

vector<string> graph[26][26];

vector<int> indegree, outdegree;

void makeGraph(const vector<string>& words)
{
    for (int i = 0; i < 26; i++)
        for (int j = 0; j < 26; j++)
            graph[i][j].clear();
    adj = vector<vector<int>>(26, vector<int>(26, 0));
    indegree = outdegree = vector<int>(26, 0);
    for (int i = 0; i < words.size(); i++) {
        int a = words[i][0] - 'a';
        int b = words[i][words[i].size() - 1] - 'a';
        graph[a][b].push_back(words[i]);
        adj[a][b]++;
        outdegree[a]++;
        indegree[b]++;
    }
}

void getEulerCircuit( int here,vector<int>& circuit )
{
    for (int there = 0; there < adj.size(); ++there) {
        while (adj[here][there] > 0) {
            adj[here][there]--;
            getEulerCircuit(there, circuit);
        }
    }

    circuit.push_back(here);
}

vector<int> getEulerTrailOrCircuit()
{
    vector<int> circuit;

    for (int i = 0; i < 26; i++) {
        if (outdegree[i] == indegree[i] + 1) {
            getEulerCircuit(i, circuit);
            return circuit;
        }
    }

    for (int i = 0; i < 26; ++i)
    {
        if (outdegree[i]) {
            getEulerCircuit(i, circuit);
            return circuit;
        }
    }

    return circuit;
}

string solve(const vector<string>& words)
{
    makeGraph(words);

    vector<int> circuit = getEulerTrailOrCircuit();

    if (circuit.size() != words.size() + 1) return "IMPOSSIBLE";

    reverse(circuit.begin(), circuit.end());
    string ret;
    for (int i = 1; i < circuit.size(); i++) {
        int a = circuit[i - 1], b = circuit[i];
        if (ret.size()) ret += " ";
        ret += graph[a][b].back();
        graph[a][b].pop_back();
    }

    return ret;
}

int n, m;
int main()
{
    cin >> n;
    while (n--) {
        cin >> m;
        vector<string> words;
        while (m--) {
            string s;
            cin >> s;
            words.push_back(s);
        }

        cout << solve(words) << endl;

    }

    return 0;
}

 先统计有向图的出入度 判断有无欧拉回路与欧拉路径后 在进行dfs的做法

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

vector<vector<int>> adj;

vector<string> graph[26][26];

vector<int> indegree, outdegree;

void makeGraph(const vector<string>& words)
{
    for (int i = 0; i < 26; i++)
        for (int j = 0; j < 26; j++)
            graph[i][j].clear();
    adj = vector<vector<int>>(26, vector<int>(26, 0));
    indegree = outdegree = vector<int>(26, 0);
    for (int i = 0; i < words.size(); i++) {
        int a = words[i][0] - 'a';
        int b = words[i][words[i].size() - 1] - 'a';
        graph[a][b].push_back(words[i]);
        adj[a][b]++;
        outdegree[a]++;
        indegree[b]++;
    }
}

void getEulerCircuit(int here, vector<int>& circuit)
{
    for (int there = 0; there < adj.size(); ++there) {
        while (adj[here][there] > 0) {
            adj[here][there]--;
            getEulerCircuit(there, circuit);
        }
    }

    circuit.push_back(here);
}


string solve(const vector<string>& words)
{
    makeGraph(words);

    int start = -1; int end = -1;

    for (int i = 0; i < 26; i++) {
        if (outdegree[i] != indegree[i]) {
            if (outdegree[i] == indegree[i] + 1 ) {
                if (-1 == start)
                    start = i;
                else
                    return "IMPOSSIBLE";
            }
            else if (outdegree[i] + 1 == indegree[i] ) {
                if( -1 == end)
                    end = i;
                else
                    return "IMPOSSIBLE";
            }
        }
    }

    vector<int> circuit;
    if (start == -1 && end == -1) {
        for (int i = 0; i < 26; ++i)
        {
            if (outdegree[i]) {
                getEulerCircuit(i, circuit);
                break;
            }
        }
    }
    else {
        getEulerCircuit(start, circuit);
    }


    reverse(circuit.begin(), circuit.end());
    string ret;
    for (int i = 1; i < circuit.size(); i++) {
        int a = circuit[i - 1], b = circuit[i];
        if (ret.size()) ret += " ";
        ret += graph[a][b].back();
        graph[a][b].pop_back();
    }

    return ret;
}

int n, m;
int main()
{
    cin >> n;
    while (n--) {
        cin >> m;
        vector<string> words;
        while (m--) {
            string s;
            cin >> s;
            words.push_back(s);
        }

        cout << solve(words) << endl;

    }

    return 0;
}

2 个人觉得 可以使用首尾单词作为关键字 去哈希 然后进行哈希查找与DFS结合的搜索 看看最后能否将单词全部使用 代码如下 

TLE

#include <iostream>
#include <vector>
#include <string>


using namespace std;

int n, m;

/*
3
4
dog
god
dragon
need
3
aa
ab
bb
2
ab
cd


need dog god dragon
aa ab bb
IMPOSSIBLE
*/

vector<string> ret;

void Dfs( int begIdx,vector<vector<vector<string>>>& vvstr)
{
    if (ret.size() == m) {
        return;
    }

    for (int i = 0; i < 26; i++) {
        for (int j = 0; j < vvstr[begIdx][i].size(); j++) {
            //选择 nextBegIdx 开头的字母
            //如果已经选择过了 则进行下一次尝试
            if (vvstr[begIdx][i][j][0] == '#') 
                continue;

            //开始选择该单词接龙
            int nextBegIdx = vvstr[begIdx][i][j].back() - 'a';
            ret.push_back(vvstr[begIdx][i][j]);
            vvstr[begIdx][i][j][0] = '#'; //修改字符串的第一个字符 作为已经被选择的标记
        
            Dfs( nextBegIdx, vvstr);

            if (ret.size() == m) {
                return;
            }

            vvstr[begIdx][i][j][0] = 'a'+ begIdx;
            ret.pop_back();
            break;
        }
    }

}


void DfsStart(vector<vector<vector<string>>>& vvstr)
{
    for (int i = 0; i < 26; i++) {
        for (int j = 0; j < 26; j++) {
            for (int k = 0; k < vvstr[i][j].size(); k++) {
                //选中这个单词作为开始
                int nextBegIdx = vvstr[i][j][k].back() - 'a';
                ret.push_back(vvstr[i][j][k]);
                vvstr[i][j][k][0] = '#'; //修改字符串的第一个字符 作为已经被选择的标记
                Dfs( nextBegIdx,vvstr);

                if (ret.size() == m) {
                    return;
                }

                //复原 继续下一次选择
                vvstr[i][j][k][0] = 'a'+i;
                ret.pop_back();
                break;
            }
        }
    }

}


int main()
{
    cin >> n;
    while (n--) {
        cin >> m;
        string s; vector<vector<vector<string>>> vvstr(26, vector<vector<string>>(26, vector<string>()));
        ret.clear();
        for (int i = 0; i < m; i++){
            cin >> s;
            int beg = s[0] - 'a';
            int end = s.back() - 'a';
            vvstr[beg][end].push_back(s);
        }
        DfsStart(vvstr);
        if(!ret.empty()){
            for (int i = 0; i < ret.size(); i++) {
                cout << ret[i] << " ";
            }
        }
        else {
            cout << "IMPOSSIBLE";
        }
        cout << endl;
    }

    return 0;
}